XPost: comp.theory, sci.logic

On 6/19/2023 3:08 PM, Ben Bacarisse wrote:

Just a reminder that you are arguing with someone who has declared that
the wrong answer is the right one:

Me: "do you still assert that [...] false is the "correct" answer even
though P(P) halts?"

PO: Yes that is the correct answer even though P(P) halts.

Because
*Ben Bacarisse targets my posts to discourage honest dialogue*
*Ben Bacarisse targets my posts to discourage honest dialogue*
*Ben Bacarisse targets my posts to discourage honest dialogue*
All of my posts will be entitled as a Rebuttal to Ben

It is an easily verified fact that P correctly simulated by H cannot
possibly reach its own last instruction and terminate normally thus from
the Professor Sipser agreed criteria the input to H(P,P) does not halt.

MIT Professor Michael Sipser has agreed that the following verbatim
words are correct (he has not agreed to anything else):
(a) If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running
unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.

To address what looks like a contradiction to reviewers not having a
very deep understanding of the halting problem:

(1) A return value of 1 from H(D,D) means the input to H(D,D) has halted

(2) A return value of 0 from H(D,D) has been redefined to mean
(a) D does not halt
(b) D has been defined to do the opposite of whatever Boolean value
that H returns.

THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL

*Now for the new material*
I am specifically defining the set of “halting problem” finite string
pair instances such that TMD2 does the opposite of the Boolean value
that each element of TMD1 returns.

The above set is the set where the behavior of the directly executed
TMD2(TMD2) is out-of-sync with the return value of TMD1(TMD2,TMD2)
*ONLY* because TMD2 has been defined to contradict whatever Boolean
value that TMD1 returns.

This makes each {TMD1, TMD2} pair isomorphic to the Liar Paradox and the
Liar Paradox: "This sentence is not true" is an unsound statement, thus
the question: Is the Liar Paradox true or false becomes an unsound
question.

Thus the question: "Does this input halt?"
is isomorphic to this question:
Is the Liar Paradox true or false?
For every {TMD1, TMD2} pair defined above.

We can know in advance that every Boolean return value from every
element of the set of TMD1 deciders necessarily out-of-sync with the
behavior of its corresponding TMD2 input because every element of the
{TMD1, TMD2} pairs has been defined to have that property.

When anyone says that we have to wait and see it is obvious that they
are only playing deceptive head games.

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic

On 6/22/2023 10:10 AM, olcott wrote:

On 6/19/2023 3:08 PM, Ben Bacarisse wrote:

Just a reminder that you are arguing with someone who has declared that the wrong answer is the right one:

Me: "do you still assert that [...] false is the "correct" answer even
      though P(P) halts?"

PO: Yes that is the correct answer even though P(P) halts.

Because
*Ben Bacarisse targets my posts to discourage honest dialogue*
*Ben Bacarisse targets my posts to discourage honest dialogue*
*Ben Bacarisse targets my posts to discourage honest dialogue*
All of my posts will be entitled as a Rebuttal to Ben

It is an easily verified fact that P correctly simulated by H cannot
possibly reach its own last instruction and terminate normally thus from
the Professor Sipser agreed criteria the input to H(P,P) does not halt.

MIT Professor Michael Sipser has agreed that the following verbatim
words are correct (he has not agreed to anything else):
(a) If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running
unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.

To address what looks like a contradiction to reviewers not having a
very deep understanding of the halting problem:

(1) A return value of 1 from H(D,D) means the input to H(D,D) has halted

(2) A return value of 0 from H(D,D) has been redefined to mean
   (a) D does not halt
   (b) D has been defined to do the opposite of whatever Boolean value
       that H returns.

THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL

*Now for the new material*
I am specifically defining the set of “halting problem” finite string pair instances such that TMD2 does the opposite of the Boolean value
that each element of TMD1 returns.

The above set is the set where the behavior of the directly executed TMD2(TMD2) is out-of-sync with the return value of TMD1(TMD2,TMD2)
*ONLY* because TMD2 has been defined to contradict whatever Boolean
value that TMD1 returns.

This makes each {TMD1, TMD2} pair isomorphic to the Liar Paradox and the
Liar Paradox: "This sentence is not true" is an unsound statement, thus
the question: Is the Liar Paradox true or false becomes an unsound
question.

Thus the question: "Does this input halt?"
is isomorphic to this question:
Is the Liar Paradox true or false?
For every {TMD1, TMD2} pair defined above.

We can know in advance that every Boolean return value from every
element of the set of TMD1 deciders necessarily out-of-sync with the
behavior of its corresponding TMD2 input because every element of the
{TMD1, TMD2} pairs has been defined to have that property.

When anyone says that we have to wait and see it is obvious that they
are only playing deceptive head games.

TMD1 is transformed from being a recognizer applied to TMD2 where TMD1
gets stuck in a loop into a decider for TMD2 that correctly determines
that TMD2 has pathological behavior relative to TMD1.


--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic

On 6/22/23 11:10 AM, olcott wrote:

On 6/19/2023 3:08 PM, Ben Bacarisse wrote:

Just a reminder that you are arguing with someone who has declared that the wrong answer is the right one:

Me: "do you still assert that [...] false is the "correct" answer even
      though P(P) halts?"

PO: Yes that is the correct answer even though P(P) halts.

Because
*Ben Bacarisse targets my posts to discourage honest dialogue*
*Ben Bacarisse targets my posts to discourage honest dialogue*
*Ben Bacarisse targets my posts to discourage honest dialogue*
All of my posts will be entitled as a Rebuttal to Ben

So, it appears that YOU don't what Honest Dialog, because you don't know
what it means.

It is an easily verified fact that P correctly simulated by H cannot
possibly reach its own last instruction and terminate normally thus from
the Professor Sipser agreed criteria the input to H(P,P) does not halt.

So, that isn't the halting problem, and your H that gives an answer
never does a "Correct Simulation" so your criteria is just invalid, as
is all your logic.

MIT Professor Michael Sipser has agreed that the following verbatim
words are correct (he has not agreed to anything else):
(a) If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running
unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.

Yes, *IF* H sees that a *CORRECT SIMULATION*, by which he means a
simulation like a UTM, which means a simulation that doesn't stop until finished, if this exact input (not one with H changed to this altered
machine, since that isn't what H is) then H could abort its simulation.
Since H DOES abort its simulation, this doesn't happen, so H has no
grounds to "correctly" abort its simulation, so it gives the wrong answer.

To address what looks like a contradiction to reviewers not having a
very deep understanding of the halting problem:

(1) A return value of 1 from H(D,D) means the input to H(D,D) has halted

(2) A return value of 0 from H(D,D) has been redefined to mean
   (a) D does not halt
   (b) D has been defined to do the opposite of whatever Boolean value
       that H returns.

So, you

THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL

No, it proves you are a LIAR, as you claim to be doing the Halting
Problem, but not using the criteria of the Halting Problem, thus all you
work is just a LIE.

*Now for the new material*
I am specifically defining the set of “halting problem” finite string pair instances such that TMD2 does the opposite of the Boolean value
that each element of TMD1 returns.

So, you are admitting you aren't doing the Halting Problem, since that
isn't the statement of the halting problem.

Again, you are just admitting that you are a liar.

The above set is the set where the behavior of the directly executed TMD2(TMD2) is out-of-sync with the return value of TMD1(TMD2,TMD2)
*ONLY* because TMD2 has been defined to contradict whatever Boolean
value that TMD1 returns.

This makes each {TMD1, TMD2} pair isomorphic to the Liar Paradox and the
Liar Paradox: "This sentence is not true" is an unsound statement, thus
the question: Is the Liar Paradox true or false becomes an unsound
question.

So? it isn't the actual Halting Problem, and you whole arguemtn is shown
to be a LIE.

Thus the question: "Does this input halt?"
is isomorphic to this question:
Is the Liar Paradox true or false?
For every {TMD1, TMD2} pair defined above.

No, it just shows that you are being a LIAR, as you are using a FALSE definition of the Halting Problem. thus ALL your work is just a big LIE.

We can know in advance that every Boolean return value from every
element of the set of TMD1 deciders necessarily out-of-sync with the
behavior of its corresponding TMD2 input because every element of the
{TMD1, TMD2} pairs has been defined to have that property.

When anyone says that we have to wait and see it is obvious that they
are only playing deceptive head games.

Since nothing you have done is based on the actual Halting Problem, you
whole arguement just becomes a LIE when you try to relate it to the
actual Halting Problem.

Face it, until you shoe that you are going to work with the actual
definitions required by the problem, nothing you do is actually correct,
and you are just proving how pitiful your skills of logic are.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)